I got values of $\pu{5.0E-4}$, $\pu{2.5E-4}$, and $183$ for the following problems. However, these seem to be inaccurate and I'm not sure what to do.

Here is the procedure:

Obtain 6 disposable test tubes. Label the test tubes 1 through 6. To each of the $\pu{6 mm}$ test tubes, add $\pu{5.0 mL}$ of $\pu{2.0E–4 M}$ $\ce{KSCN}$. If you make a mistake, you must rinse out and thoroughly dry the test tube before starting over.

To test tube 1, add $\pu{5.0 mL}$ of $\pu{0.20 M}\; \ce{Fe(NO3)3}$ and stir. The resulting solution is your standard solution of $\ce{FeSCN^2+}$.

For the other test tubes, do the following: Measure $\pu{10.0 mL}$ of $\pu{0.20 M}$ $\ce{Fe(NO3)3}$ in a $\pu{25 mL}$ graduated cylinder, fill to $\pu{25.0 mL}$ with distilled water, and stir thoroughly to mix. Measure $\pu{5.0 mL}$ of the resulting $\pu{0.080 M}$ $\ce{Fe^3+}$ into test tube 2.

Discard all but $\pu{10.0 mL}$ of the $\pu{0.080 M}$ $\ce{Fe^3+}$, refill the graduated cylinder with $\pu{10.0 mL}$ of the $\pu{0.080 M}$ $\ce{Fe^3+}$ solution to $\pu{25.0 mL}$ with distilled water, and stir thoroughly. Add $\pu{5.0 mL}$ of the resulting $\pu{0.032 M}$ $\ce{Fe^3+}$ to test tube 3.

Again discard all but $\pu{10.0 mL}$ of the contents of the graduated cylinder, refill to $\pu{25.0 mL}$ with distilled water, stir, and add $\pu{5.0 mL}$ to test tube 4.

Repeat above procedure with test tubes 5 and 6.

  1. Calculate the initial concentrations of $\ce{Fe^3+}$ and $\ce{SCN-}$ for the six solutions to be prepared in this experiment (i.e., the 6 solutions prepared in test tubes 1–6). These are the concentrations just after the two solutions are mixed in the large test tube (thereby diluting each other) but before any $\ce{FeSCN^2+}$ has formed. Assume that the volumes are additive and that the $\ce{Fe(NO3)3}$ and $\ce{KSCN}$ are each completely dissociated. What is the concentration of $\ce{Fe^3+}$ in test tube #5?

$$ \begin{array}{c|cc} \hline \text{Test tube} & \text{Initial}~[\ce{Fe^3+}] & \text{Initial}~[\ce{SCN-}] \\ \hline 1 & & \\ 2 & & \\ 3 & & \\ 4 & & \\ 5 & & \\ 6 & & \\ \hline \end{array} $$

  1. For the chart in question 1, what is the molar concentration of $\ce{SCN-}$ in tube #4?

  2. If the equilibrium $\ce{[FeSCN^2+]}$ in test tube 2 was determined to be $\pu{8.3E-5 M}$, calculate the equilibrium $\ce{[Fe^3+]}$.


You must first calculate dilution factors. Assuming the volumes are additive, we can calculate dilution using $M_1V_1=M_2V_2$ equation. Also, assuming $\ce{Fe(NO3)3}$ and $\ce{KSCN}$ are each completely dissociated, we can say:

$$\ce{[Fe(NO3)3] = [Fe^3+]} \text{ and } \ce{[KSCN] = [SCN-]}$$

Thus, initial $\ce{[Fe^3+]}$ and $\ce{[SCN-]}$ are $0.20$ and $\pu{2.0 \times 10^{-4} M}$, respectively.

In test tube one: $\pu{5.0 mL}$ of $\pu{0.20 M} \; \ce{Fe(NO3)3}$ and $\pu{5.0 mL}$ of $\pu{2.0 × 10^{−4} M} \; \ce{KSCN}$ are added together, thus dilution factor is 1:2. Therefore, using $M_1V_1=M_2V_2$ equation you can calculate initial concentrations of $\ce{[Fe^3+]}$ and $\ce{[SCN-]}$ after the addition (but before they react with each other) as follows:

$$\ce{[Fe^3+]} = \pu{\frac{0.2 \times 5.0}{10.0} M} = \pu{0.10 M} \text{ and } \ce{[SCN-]} = \pu{\frac{2.0 × 10^{−4} \times 5.0}{10.0} M} = \pu{1.0 × 10^{−4} M}$$

As a matter of fact, since the original solution of $\ce{KSCN}$ did not undergo any dilution maneuver before final 1:1 dilution, $\ce{[SCN-]}$ of all 6 test tubes are same, and its value is $\pu{1.0 × 10^{−4} M}$.

On the other hand, the original solution of $\ce{Fe(NO3)3}$ has undergone serial dilution of $\pu{10 mL}$ to $\pu{25 mL}$ (hence dilution factor is $10/25$ or $2/5$). Therefore, if original concentration of each solution is $x$ before add together in test tubes, the final $\ce{[Fe^3+]}_a$ before add with $\ce{[SCN-]}$ is:

$$\ce{[Fe^3+]}_a = \frac{x \times 2.0}{5.0} \: \pu{M}$$

Thus, the final $\ce{[Fe^3+]}_a$ before add with $\ce{[SCN-]}$:

$$\ce{[Fe^3+]}_2 = \pu{\frac{0.20 \times 2.0}{5.0} M}= \pu{0.08 M}, \, \ce{[Fe^3+]}_3 = \pu{\frac{0.08 \times 2.0}{5.0} M}= \pu{0.032 M}, \, \ce{[Fe^3+]}_4 = \pu{\frac{0.032 \times 2.0}{5.0} M}= \pu{0.0128 M}, \, \ce{[Fe^3+]}_5 = \pu{\frac{0.0128 \times 2.0}{5.0} M}= \pu{0.00512 M}, \, \ce{[Fe^3+]}_6 = \pu{\frac{0.00512 \times 2.0}{5.0} M}= \pu{0.002048 M}$$

Finally, those concentrations were further diluted in 1:1 ratio in test tubes, and therefore, initial $\ce{[Fe^3+]}$ are half of those $\ce{[Fe^3+]}_a$ values. I tabulated all of them accordingly in following table, as answer to your question 1:

$$ \begin{array}{c|cc} \text{Test Tube #} &\text{Initial $\ce{[Fe^3+]}$} &\text{Initial $\ce{[SCN-]}$} \\\hline 1 & \pu{0.10 M} & \pu{1.0 × 10^{−4} M}\\ 2 & \pu{0.04 M} & \pu{1.0 × 10^{−4} M}\\ 3 & \pu{0.016 M} & \pu{1.0 × 10^{−4} M}\\ 4 & \pu{0.0064 M} & \pu{1.0 × 10^{−4} M}\\ 5 & \pu{0.00256 M} & \pu{1.0 × 10^{−4} M}\\ 6 & \pu{0.001024 M} & \pu{1.0 × 10^{−4} M}\\\hline \end{array} $$

  1. There are 7 different values in the table. But you have given one relevant answer, and it is also incorrect.
  2. Find the answer for $\ce{[SCN-]}$ in test tube 4 from the table, which is identical in all 6 test tubes.
  3. The concern equilibrium is: $$\ce{Fe^3+ + SCN- <=> FeSCN^2+}$$ Reaction mole ratio of $\ce{[Fe^3+]}$ : $\ce{[SCN-]}$ : $\ce{[FeSCN^2+]}$ is 1 : 1 : 1. If equilibrium concentration of $\ce{[FeSCN^2+]}$ in test tube 2 is $\ce{ 8.3×10^{−5} M}$, the equilibrium concentration of $\ce{[Fe^3+]}$ in test tube 2 is $\ce{ (0.04 - 8.3×10^{−5}) M}$ and that of $\ce{[SCN-]}$ in test tube 2 is $\ce{ (1.0×10^{−4} - 8.3×10^{−5}) M}$. Now, you may able to calculate equilibrium constent for the reaction using following equation:

$$K_\mathrm{eq} = \frac{\ce{[FeSCN^2+]}}{\ce{[Fe^3+]}\ce{[SCN-]}}$$

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    $\begingroup$ @Denise If you find the answer useful, the best way to appreciate the effort that went into producing one would be to upvote and accept the answer. $\endgroup$ – andselisk Oct 22 '19 at 5:57

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